# How to reflect quantum uncertainty in statistical mechanics?

How does the superposition principle of the quantum mechanics reconcile with the statistical mechanics? Because in deriving canonical distribution, I don't see anything related to superposition principle, but canonical distribution can be used to describe quantum systems, for example, a single quantum oscillator in thermal contact with a heat reservoir.So how to derive canonical distribution not from the eigenstates of the Hamiltonian but from the superposition states?

More generally speaking, how to reflect quantum uncertainty in statistical mechanics?

• In large-scale systems such as a bunch of molecules in a gas or a population of people, one often finds the micro particles to be identical. In this case, physicists take a mean-field view in which they model the interactive behavior of the individual micro components as aggregated in order to make the mathematics scalable. Does the question refer to this approach or does it expect examples of applications? Mar 1, 2020 at 10:05
• @kb314 Thanks! Can you give more explanation and some examples? Mar 1, 2020 at 10:10
• You're welcome! If you are familiar with statistical mechanics I would suggest reading about the Fokker-Planck equation which formalizes the mean-field point of view for general random processes. On the other hand, I found it very useful to read this article which explains in detail how this concept of the micro-macro interaction and averaging them to arrive at a tractable theory is arrived at using the mean-field idea, for various large-scale stochastic systems. Mar 1, 2020 at 10:15
• Are you familiar with the density matrix formalism? Are you asking how to construct quantum statistical mechanics ensembles, or clarifications on specific aspect of them?
– fqq
Mar 3, 2020 at 12:47
• Many statmech/condmat books have introductions to density matrices and quantum statistical mechanics. You should probably start with those (depending on your taste), or even take a course about it if you have access to it (it is often taught in courses on advacend QM, statmech, condmat...). It is a vast and important subject.
– fqq
Mar 3, 2020 at 13:13

There are many cases that you can use statistical mechanics for various reasons. Regarding your question about the ensemble, it is possible to use them even in a single system.

Let's consider the most familiar example, the ideal gas, in which various physical properties have be derived. Now, let's divide the system with volume $$V$$ into $$N$$ sub-system with volume $$V/N$$. The $$N$$ should not be too large such that $$V/N$$ can be considered as microscopically large. Hence, all of the sub-system have the similar (mean) macroscopic properties as the originally system, but now you have $$N$$ realization of an ensemble instead of a single system! Even for interacting system, it is usually possible to divide them into multiple sub-system, if the division is larger than correlation length. Each sub-system can now be considered as a realization of the ensemble, and statistics can be performed for this ensemble.

On the other hand, if the system has the ergodic property. Then the system at different time may be considered as a realization of the ensemble, with sufficiently long time separated. The time averaged properties will be the same as the ensemble averaged. As an example, the temperature may fluctuate over very short time period, but a slow response thermometer can report the average temperature.

• But in explaining paramagnetism through canonical ensemble consideration of an individual atom in a lattice site of a solid, all the other atoms and all the other degrees of freedom are regarded as consitituting a heat reservoir. So in case of the statistical interpretation of paramagnetism how is the ensemble constructed? Mar 1, 2020 at 13:05
• And in considering the velocity distribution of an ideal gas, also a single molecule or atom is considered as a microsystem in contact with the other part of the gas which serves as a heat reservoir. And applying canonical distribution to this microsystem we obtain the velocity distribution. In this case, how is the ensemble constructed? Mar 1, 2020 at 13:09
• In my opinion, ensemble is purely an imaginary object with large number of accessible states used to calculate properties of a system with large number of degrees of freedom. Mar 1, 2020 at 14:12

The answer to your question "how to put quantum uncertainty into statistical mechanics" is: with the density matrix.

Full description is in textbooks etc,(Feynman's Lectures on Statistical Mechanics, and von Neumann's Mathematical Foundations of Quantum Mechanics are good) but briefly you have a standard ensemble (canonical, microcanonical, or whatever) of $$N$$ systems with system $$i$$ in state $$|\phi_i>$$ (not necessarily eigenstates).

Introduce the density operator, $$\hat \rho={1 \over N} \sum_{i=1}^N |\phi_i> <\phi_i|$$.

That's the statistical mechanics bit. Now the quantum. Introduce a basis set of states $$|\psi_j>$$ which are eigenstates of some operator you're interested in (you have a choice), and the density matrix is defined using these states. $$\rho_{jk}=<\psi_j| \hat \rho|\psi_k>={1 \over N} \sum_{i=1}^N<\psi_j |\phi_i> <\phi_i|\psi_k>$$.

The diagonal elements $$\rho_{kk}$$ then give the probability that if you measure a random member of the ensemble it will turn our to be in the $$k$$ eigenstate, and that includes the statistical probability, from the $${ 1\over N } \sum$$, and the quantum probability from the $$|<\psi|\phi>|^2$$. It can also be used to give expectation values for operators.

The off-diagonal elements contain information about the nature of the uncertainty. A classic example is an ensemble of electrons which are half spin up and half spin down, compared to an ensemble which are all spin sideways (aligned along, say, the x axis). In both cases, both diagonal elements are just $$1 \over 2$$, whereas the off diagonals are zero in the first case but not the second.

Incidentally, the whole 'collapse of the wave function' business is equivalent to setting all off diagonal elements of the density matrix to zero.

We can use statistical mechanics whenever the phase space distribution of the system is independent of time. This means that the distribution has to be a function of the constants of motion of the system which are $$E, \vec L, \vec p$$. Generally we have control over $$E$$ and not the others so we generally deal with systems whose phase space distribution is purely a function if $$E$$.

Now the way we go about applying this is by first working out the allowed energy level for a single particle. Here we consider the total Hamiltonian (potential) that a particle sees and then find out the allowed energy states. Note that this is not the same as taking a system of particles and choosing one of them and treating rest as heat bath. No no. We are solving for possible energy levels for a single particle. Once we have that, we will fill them up with the total number of particles such that the total energy is equal to the desired value.

And there could be many ways of filling it such that we get the right total energy. These are all the possible microstates. This method works when the interaction between particles is negligible, because we ignore interactions when we solve for single particle states.

• You say this is not the same as taking a system of particles and choosing one of them and treating rest as heat bath. But in Frederick Reif's book he indeed does so when treating paramagnetism and velocity distribution of an ideal gas. Mar 2, 2020 at 3:33
• We consider the case of a small system A in thermal interaction with a heat reservoir A'. The system A may be any relatively small macroscopic system. (For example, it may be a bottle of wine immersed in a swimming pool, the pool acting as a heat reservoir.) Sometimes it may also be a distinguishable macroscopic system which can be clearly identified.(For example, it may be an atom at some lattice site in a solid, the solid acting as a heat reservoir.) Mar 2, 2020 at 3:36
• I see. Your opinion is partly right and partly wrong. The right part is according to your way we can obtain the right answer. The wrong part is that actually, for distinguishable particles, taking a system of particles and choosing one of them and treating rest as heat bath is the same as filling single particle levels up with the total number of particles such that the total energy is equal to the desired value. Mar 2, 2020 at 8:00
• As a test, you can use both ways to obtain Curie' law for paramagnetism. Mar 2, 2020 at 8:01